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- anonymous

let T,U be two linear operators on a vector space V, Show that TU and UT have the same eigen values.

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- anonymous

let T,U be two linear operators on a vector space V, Show that TU and UT have the same eigen values.

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- anonymous

I'll get you started.
Suppose that \[\lambda\] is a non-zero eigenvalue for TU. Then
\[
TU v -\lambda v = 0
\]
for some non-zero vector v. So, applying U to both sides,
\[
UT (Uv) -\lambda U v = 0.
\]
So long as Uv is not zero this means that \[\lambda\] is an eigenvalue for UT. But Uv can't be zero because then TUv would be zero, so \[\lambda\] would have been zero to begin with.
The case of zero eigenvalue can be done similarly. Does anybody know how to make an equation appear inline rather than on its own line? I'm new here.

- anonymous

how about this let Tv = av and Uv = bv a and b are the respective eigen values.
TUv = Tbv = bTv = bav
UTv = Uav = aUv=abv=bav so TUv =UTv =bav so UT and TU must have the same eigen values.

- anonymous

No, you've assumed that T and U have the same eigenvector v. This is not part of the hypothesis and might not be true in general. You must start with the assumption that you have an eigenvalue for TU (or UT) then show that this is an eigenvalue for UT (or TU), with no particular assumptions about the eigenvector. You see in my explanation above that we end up with Uv an eigenvector for UT when we start with v an eigenvector for TU, both with the same eigenvalue.

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- anonymous

but the eigen values a and b are not the same

- anonymous

Yes, but you've assumed that Tv =av and Uv = bv, i.e. that v is a common eigenvector for T and U, even though there are different eigenvalues. A common eigenvalue does not always exist, and even if it did it would not solve the original problem.
Start with an eigenvalue for TU, then show that it is an eigenvalue for UT (in case with different eigenvectors). See if you can follow the argument in my first post.

- anonymous

cool. so we must assume that the eignevectors are not the same too thanx.

- anonymous

No assumptions on the eigenvectors. Assume a is an eigenvalue for TU, show that a is an eigenvalue for UT.

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